$(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ is said to have a general solution of $x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$.
My Approach:
Considering the equation as
$$
a\cos x+b\sin x=\sqrt{a^2+b^2}\Big(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\Big)=\sqrt{a^2+b^2}\big(\sin y.\cos x+\cos y.\sin x\big)=\sqrt{a^2+b^2}.\sin(y+x)=2
$$
$\frac{a}{\sqrt{a^2+b^2}}=\sin y$ and $\frac{b}{\sqrt{a^2+b^2}}=\cos y$.
$$
{\sqrt{a^2+b^2}}=\sqrt{8}=2\sqrt{2}\\\tan y=a/b=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{2}}-\frac{1}{2}.\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{2}}+\frac{1}{2}.\frac{1}{\sqrt{2}}}=\frac{\sin(\pi/3-\pi/4)}{\sin(\pi/3+\pi/4)}=\frac{\sin(\pi/3-\pi/4)}{\cos(\pi/3-\pi/4)}=\tan(\pi/3-\pi/4)\implies y=\pi/3-\pi/4=\pi/12
$$
Substituting for $y$,
$$
2\sqrt{2}.\sin(\frac{\pi}{12}+x)=2\implies \sin(\frac{\pi}{12}+x)=\frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\\\implies \frac{\pi}{12}+x=n\pi+(-1)^n\frac{\pi}{4}\implies x=n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{12}
$$
What's going wrong with the approach ?
Best Answer
Our hint is: $a\cos \theta +b\sin \theta =c$.
Given: $(\sqrt{3}-1)\cos \theta +(\sqrt{3}+1)\sin \theta =2$.
Let $(\sqrt{3}-1) = r\cos \alpha$ and $(\sqrt{3}+1) =r\sin \alpha$.
Then $r\cos \alpha \cos \theta + r\sin \alpha \sin \theta =2 \Rightarrow r\cos(\theta-\alpha) =2 \Rightarrow \cos(\theta-\alpha) =\frac{2}{r}$.
Now, $r =\sqrt{(\sqrt{3}-1)^2 +(\sqrt{3}+1)^2} = \sqrt{8} =2\sqrt{2}$.
Thus, $\cos(\theta-\alpha) =\frac{1}{\sqrt{2}} = \cos \frac{\pi}{4}$.
Also, $\tan \alpha =\frac{\sqrt{3}+1}{\sqrt{3}-1} = \tan(\frac{\pi}{2}-\frac{\pi}{3} +\frac{\pi}{4}) \Rightarrow \alpha =\frac{5\pi}{12}$.
Thus:$(\theta-\alpha) =2n\pi \pm \frac{\pi}{4}$. Giving, $\theta = 2n\pi \pm \frac{\pi}{4} +\frac{5\pi}{12}.$